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4 years ago

14

Gravitational force of attraction "F" exists between two point masses A and B when a fixed distance separates them. After mass A

is tripled and mass B is halved, the gravitational attraction between the two masses is

1 answer:

Umnica [9.8K]4 years ago

8 0

Answer:

1.5 F

Explanation:

F = G m1 m2/ r^2

new force F' = G × 3m1 × 0.5 m2 / r^2

F' = 1.5 ×G m1 m2/ r^2

F' = 1.5 F

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(a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

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Given that,

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Voltage $v=3\cos4\pi t\ V$

Time t = 0.3 sec

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Using formula of current

$i(t)=\dfrac{dq}{dt}$

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$i(t)=5\times4\pi\cos4\pi t$

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(a).We need to calculate the power delivered to the element

Using formula of power

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Put the value into the formula

$p(t)=3\cos4\pi t\times20\pi\cos4\pi t$

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$p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi t}{2})$

Put the value of t

$p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi\times0.3}{2})$

$p(t)=30\pi\times10^{-3}(1+\cos8\pi \times0.3)$

$p(t)=187.68\ mW$

(b). We need to calculate the energy delivered to the element between 0 and 0.6 s

Using formula of energy

$E(t)=\int_{0}^{t}{p(t)dt}$

Put the value into the formula

$E(t)=\int_{0}^{0.6}{30\pi\times10^{-3}(1+\cos8\pi \times t)}$

$E(t)=30\pi\times10^{-3}\int_{0}^{0.6}{1+\cos8\pi \times t}$

$E(t)=30\pi\times10^{-3}(t+\dfrac{\sin8\pi t}{8\pi})_{0}^{0.6}$

$E(t)=30\pi\times10^{-3}(0.6+\dfrac{\sin8\pi\times0.6}{8\pi}-0-0)$

$E(t)=57.52\ mJ$

Hence, (a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

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